LeetCode #1275 — EASY

Find Winner on a Tic Tac Toe Game

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Tic-tac-toe is played by two players A and B on a 3 x 3 grid. The rules of Tic-Tac-Toe are:

Players take turns placing characters into empty squares ' '.
The first player A always places 'X' characters, while the second player B always places 'O' characters.
'X' and 'O' characters are always placed into empty squares, never on filled ones.
The game ends when there are three of the same (non-empty) character filling any row, column, or diagonal.
The game also ends if all squares are non-empty.
No more moves can be played if the game is over.

Given a 2D integer array moves where moves[i] = [row_i, col_i] indicates that the i^th move will be played on grid[row_i][col_i]. return the winner of the game if it exists (A or B). In case the game ends in a draw return "Draw". If there are still movements to play return "Pending".

You can assume that moves is valid (i.e., it follows the rules of Tic-Tac-Toe), the grid is initially empty, and A will play first.

Example 1:

Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
Output: "A"
Explanation: A wins, they always play first.

Example 2:

Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: B wins.

Example 3:

Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
Output: "Draw"
Explanation: The game ends in a draw since there are no moves to make.

Constraints:

1 <= moves.length <= 9
moves[i].length == 2
0 <= row_i, col_i <= 2
There are no repeated elements on moves.
moves follow the rules of tic tac toe.

Step 01

Brute Force Baseline

Problem summary: Tic-tac-toe is played by two players A and B on a 3 x 3 grid. The rules of Tic-Tac-Toe are: Players take turns placing characters into empty squares ' '. The first player A always places 'X' characters, while the second player B always places 'O' characters. 'X' and 'O' characters are always placed into empty squares, never on filled ones. The game ends when there are three of the same (non-empty) character filling any row, column, or diagonal. The game also ends if all squares are non-empty. No more moves can be played if the game is over. Given a 2D integer array moves where moves[i] = [rowi, coli] indicates that the ith move will be played on grid[rowi][coli]. return the winner of the game if it exists (A or B). In case the game ends in a draw return "Draw". If there are still movements to play return "Pending". You can assume that moves is valid (i.e., it follows the rules of

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[[0,0],[2,0],[1,1],[2,1],[2,2]]

Example 2

[[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]

Example 3

[[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]

Core Insight

What unlocks the optimal approach

It's straightforward to check if A or B won or not, check for each row/column/diag if all the three are the same.
Then if no one wins, the game is a draw iff the board is full, i.e. moves.length = 9 otherwise is pending.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1275: Find Winner on a Tic Tac Toe Game
class Solution {
    public String tictactoe(int[][] moves) {
        int n = moves.length;
        int[] cnt = new int[8];
        for (int k = n - 1; k >= 0; k -= 2) {
            int i = moves[k][0], j = moves[k][1];
            cnt[i]++;
            cnt[j + 3]++;
            if (i == j) {
                cnt[6]++;
            }
            if (i + j == 2) {
                cnt[7]++;
            }
            if (cnt[i] == 3 || cnt[j + 3] == 3 || cnt[6] == 3 || cnt[7] == 3) {
                return k % 2 == 0 ? "A" : "B";
            }
        }
        return n == 9 ? "Draw" : "Pending";
    }
}

// Accepted solution for LeetCode #1275: Find Winner on a Tic Tac Toe Game
func tictactoe(moves [][]int) string {
	n := len(moves)
	cnt := [8]int{}
	for k := n - 1; k >= 0; k -= 2 {
		i, j := moves[k][0], moves[k][1]
		cnt[i]++
		cnt[j+3]++
		if i == j {
			cnt[6]++
		}
		if i+j == 2 {
			cnt[7]++
		}
		if cnt[i] == 3 || cnt[j+3] == 3 || cnt[6] == 3 || cnt[7] == 3 {
			if k%2 == 0 {
				return "A"
			}
			return "B"
		}
	}
	if n == 9 {
		return "Draw"
	}
	return "Pending"
}

# Accepted solution for LeetCode #1275: Find Winner on a Tic Tac Toe Game
class Solution:
    def tictactoe(self, moves: List[List[int]]) -> str:
        n = len(moves)
        cnt = [0] * 8
        for k in range(n - 1, -1, -2):
            i, j = moves[k]
            cnt[i] += 1
            cnt[j + 3] += 1
            if i == j:
                cnt[6] += 1
            if i + j == 2:
                cnt[7] += 1
            if any(v == 3 for v in cnt):
                return "B" if k & 1 else "A"
        return "Draw" if n == 9 else "Pending"

// Accepted solution for LeetCode #1275: Find Winner on a Tic Tac Toe Game
struct Solution;

impl Solution {
    fn tictactoe(moves: Vec<Vec<i32>>) -> String {
        let mut board = [[' '; 3]; 3];
        let mut player = 'X';
        let n = moves.len();
        for m in moves {
            let r = m[0] as usize;
            let c = m[1] as usize;
            board[r][c] = player;
            player = if player == 'X' { 'O' } else { 'X' };
        }
        let ss = [
            [(0, 0), (0, 1), (0, 2)],
            [(1, 0), (1, 1), (1, 2)],
            [(2, 0), (2, 1), (2, 2)],
            [(0, 0), (1, 0), (2, 0)],
            [(0, 1), (1, 1), (2, 1)],
            [(0, 2), (1, 2), (2, 2)],
            [(0, 0), (1, 1), (2, 2)],
            [(0, 2), (1, 1), (2, 0)],
        ];
        for s in &ss {
            let mut a = 0;
            let mut b = 0;
            for p in s {
                let i = p.0;
                let j = p.1;
                match board[i][j] {
                    'X' => a += 1,
                    'O' => b += 1,
                    _ => {}
                }
            }
            if a == 3 {
                return "A".to_string();
            }
            if b == 3 {
                return "B".to_string();
            }
        }
        if n == 9 {
            return "Draw".to_string();
        }
        "Pending".to_string()
    }
}

#[test]
fn test() {
    let moves = vec_vec_i32![[0, 0], [2, 0], [1, 1], [2, 1], [2, 2]];
    let res = "A".to_string();
    assert_eq!(Solution::tictactoe(moves), res);
    let moves = vec_vec_i32![[0, 0], [1, 1], [0, 1], [0, 2], [1, 0], [2, 0]];
    let res = "B".to_string();
    assert_eq!(Solution::tictactoe(moves), res);
    let moves = vec_vec_i32![
        [0, 0],
        [1, 1],
        [2, 0],
        [1, 0],
        [1, 2],
        [2, 1],
        [0, 1],
        [0, 2],
        [2, 2]
    ];
    let res = "Draw".to_string();
    assert_eq!(Solution::tictactoe(moves), res);
    let moves = vec_vec_i32![[0, 0], [1, 1]];
    let res = "Pending".to_string();
    assert_eq!(Solution::tictactoe(moves), res);
}

// Accepted solution for LeetCode #1275: Find Winner on a Tic Tac Toe Game
function tictactoe(moves: number[][]): string {
    const n = moves.length;
    const cnt = new Array(8).fill(0);
    for (let k = n - 1; k >= 0; k -= 2) {
        const [i, j] = moves[k];
        cnt[i]++;
        cnt[j + 3]++;
        if (i == j) {
            cnt[6]++;
        }
        if (i + j == 2) {
            cnt[7]++;
        }
        if (cnt[i] == 3 || cnt[j + 3] == 3 || cnt[6] == 3 || cnt[7] == 3) {
            return k % 2 == 0 ? 'A' : 'B';
        }
    }
    return n == 9 ? 'Draw' : 'Pending';
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.